A285770 - OEIS

A285770

Let q_1<q_2<... be primes not dividing 2n+1 such that 2n-q_i + 1 is twice a prime. a(n) is smallest i such that 2n+2q_i+1 is prime or a(n)=0 if there is no such i.

0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 4, 5, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 3, 3, 1, 2, 3, 1, 1, 2, 0, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 3, 2, 1, 5 (list; graph; refs; listen; history; text; internal format)

	OFFSET	`1,11`
	COMMENTS	By the Lemoine-Levy conjecture, for every n>=3, there are primes p and q such that 2n+1=2p+q. In A277688 were considered the numbers of the form 2n+1+2p, in this sequence we consider the numbers of the form 2n+1+2q. Then to the same condition as in A277688 satisfies an extremely rare set of numbers that contains numbers {2n+1} for which a(n)=0: {11, 59, 151,...}. Comparing this with our conjecture in A277688, we conjecture here that this set is finite. For the explanation of this conjecture we need not refer to the minimal number of the representation 2n+1 for large n (most likely, it is, as in the Goldbach presentations, for 2n, more than cn/(log n)^2 with some constant c) since we have a prime only among the first several of these representations, as in this sequence. This leaves us with an important question: why does A277688 contain much more terms than the zeros in this sequence? `The positions of zeros are {1,2,5,29,75} up to 100000. - Peter J. C. Moses, Apr 26 2017` `The answer on the question follows from the following arguments. Note that if, for a prime p, 2n+1-2p is prime, then it is larger than 3 (if not, then p>=n and 2n+1-2p<=1). Now if 2n+1 is not divisible by 3 and 2n+1-6 is composite, then either 2n+1-2p or 2n+1+2p is divisible by 3 and, since the first number is prime >3, then 2n+1+2p is divisible by 3 and thus such 2n+1 is in A277688. - Vladimir Shevelev, Apr 28 2017` `No other zeros up to 5*10^6. - Michel Marcus, Apr 29 2017`
	LINKS	`Table of n, a(n) for n=1..89.`
	PROG	`(PARI) a(n) = {i = 0; forprime(p=2, 2n+1, if ((gcd(p, 2n+1)==1) && ((2n-p) % 2) && isprime((2n-p+1)/2), i++; if (isprime(2n+2p+1), return(i)); ); ); return(0); } \\ Michel Marcus, Apr 29 2017`
	CROSSREFS	`Cf. A277688.` `Sequence in context: A324049 A194510 A331311 * A370892 A174820 A099501` `Adjacent sequences: A285767 A285768 A285769 * A285771 A285772 A285773`
	KEYWORD	`nonn`
	AUTHOR	`Vladimir Shevelev, Apr 25 2017`
	EXTENSIONS	`More terms from Peter J. C. Moses, Apr 25 2017`
	STATUS	`approved`